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Let $A$ and $B$ unital $C^\ast$-algebras, $f:A\to B$ a linear, bounded map such that $f(a^*)=f(a)^*$ for all $a\in A$, $f(1_A)=1_B$ and $f(a)f(b)=0$ for all $a,b\in A_{sa}$ with $ab=0$. Follows $f(a^2)=f(a)^2$ for all $a\in A$?

I have tried to proof this first for self-adjoint elements $a\in A$ using the continuous functional calculus but I'm stuck. Do you know a proof or a reference? Greetings

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  • $\begingroup$ It's easy to see that if the $p_j$ are mutually orthogonal projections ($p=p^*$, $p_jp_k=\delta_{jk} p_j$), then so are the $f(p_j)$. This gives the claim for self-adjoint $a$. I don't quite see how this helps you with the general case, though. $\endgroup$
    – Christian Remling
    Aug 2, 2015 at 17:40
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    $\begingroup$ @ChristianRemling How does that give you self-adjoint $a$? There are unital $C^*$-algebras with no nontrivial projections (e.g. $C([0,1])$). $\endgroup$
    – Robert Israel
    Aug 2, 2015 at 18:09
  • $\begingroup$ @RobertIsrael: Well, then I guess it doesn't. I still think there's a formally correct version of the argument, though, but anyway, as I said, I wouldn't know what to do from there. $\endgroup$
    – Christian Remling
    Aug 2, 2015 at 18:18
  • $\begingroup$ This should follow from Wolff's structure theorem for disjointness preserving maps (Theorem 2.3 in Disjointness preserving operators on C*-algebras, Arch. Math. 1994 ). Your unital condition forces the map to be a Jordan morphism from which the $f(a^2)=f(a)^2$ should easily follow. $\endgroup$
    – Caleb Eckhardt
    Aug 3, 2015 at 16:02
  • $\begingroup$ Thank you guys. @Caleb Eckhard if I see correctly, Wolff proved this what I want to know in his paper. Thanks! $\endgroup$
    – Sabrina Gemsa
    Aug 5, 2015 at 17:34

2 Answers 2

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Yemon Choi pointed out in a comment to this related question that we can conclude $f$ is a *-homomorphism under the stronger assumption that $ab = 0$ implies $f(a)f(b) = 0$ for arbitrary (not just self-adjoint) $a,b \in A$. This folows from a result stated in the abstract of the paper Alaminos, Brešar, Extremera, and Villena, Maps preserving zero products, Studia Math. 193 (2009), 131-159 by taking $\phi(a,b) = f(a)f(b)$ and setting $b = 1$ in the conclusion.

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We can reduce to the case where $A$ and $B$ are commutative subalgebras, e.g. taking the closed star-algebra generated by $a$. Then $A = C(X)$ and $B = C(Y)$ for compact Hausdorff spaces $X$ and $Y$, and we have a continuous linear map $f : A \to B$. From this we get a continuous map $f^*$ between the duals $C(Y)^*$ and $C(X)^*$ in the weak* topology.

We also have natural embeddings $X \to C(X)^*$ and $Y \to C(Y)^*$ coming from point evaluations (i.e. delta-masses, in terms of measures). You can recognize the measures supported at a single point by the property that, if $\phi(ab) = 0$, then $\phi(a)=0$ or $\phi(b)=0$. By the condition $f(a)f(b)=0$ whenever $ab=0$, it follows that $f^*$ sends delta masses to multiples of delta masses; and from $f(1)=1$ we get that $f^*$ sends $Y$ continuously to $X$.

Thus $f: A \to B$ is a morphism of star-algebras, induced by a continuous map from $Y$ to $X$. In particular $f(a^2) = f(a)^2$.

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  • $\begingroup$ Why can you assume $B = C(Y)$? $\endgroup$
    – Nik Weaver
    Aug 2, 2015 at 22:51
  • $\begingroup$ You can take B to be the closure of f(A). It will then be a commutative C^* algebra and hence = C(Y) where Y is the space of maximal ideals in B. $\endgroup$
    – Curtis McMullen
    Aug 3, 2015 at 2:21
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    $\begingroup$ I don't see how we know that the closure of $f(A)$ is commutative --- $f$ isn't assumed to be a homomorphism. $\endgroup$
    – Nik Weaver
    Aug 3, 2015 at 2:23
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    $\begingroup$ Also, the result is claimed for all $a \in A$, not just self-adjoint $a$. So the C*-algebra generated by $a$ need not be commutative either. $\endgroup$
    – Nik Weaver
    Aug 3, 2015 at 2:37
  • $\begingroup$ That is a good point. $\endgroup$
    – Curtis McMullen
    Aug 3, 2015 at 3:44

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